In the picture below suppose that ABEC and AEBD Prove that A

In the picture below, suppose that AB||EC and AE||BD. Prove that [AGFH] = [DEG] + [BCH]

Solution

It is given that AB II EC and AE II BD

we also know that triangles on the same base and between same parallels are equal in area So,

DEF and AGH are equal i.e. DEF = AGF -------(1)

and AEG and BCF are also equal i.e. AEG = BCF ------(2)

Now substracting (2) from (1) then adding we will get

[DEG + BCH] = [AGFH]